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Homogeneous Systems
Henry Hermes, Matthias Kawski, Fabio Ancona
To cite this version:
HOMOGENEOUS RESONANCE & ASYMPTOTIC STABILITY FOR HOMOGENEOUS SYSTEMS
HENRY HERMES
Department of Mathematics, Box 395 University of Colorado
Boulder, CO 80309–0395, USA hermes@euclid.colorado.edu
MATTHIAS KAWSKI†
School of Mathematical and Statistical Sciences Arizona State University
Tempe, AZ 85287–1804, USA kawski@asu.edu
FABIO ANCONA Dipartimento di Matematica Universit`a degli Studi di Padova Via Trieste n.63, 35121 Padova, Italy
ancona@math.unipd.it
ABSTRACT. This paper surveys existing necessary conditions, and gives new condi-tions based on homogeneous resonance, for a homogeneous system to admit a homoge-neous (of correct order) continuous, asymptotically stabilizing, state, feedback control. Such conditions are basic in utilizing high order, homogeneous, approximations of
non-†This research was supported by the National Science Foundation grant DMS 09–08204
The third author was partially supported by the European Union 7th Framework Programme [FP7-PEOPLE-2010-ITN] under grant agreement n.264735-SADCO and by Fondazione CaRiPaRo Project ”Nonlinear Partial Differential Equations: models, analysis, and control-theoretic problems”
linear systems to construct asymptotically stabilizing feedback controls when linear approximations of the nonlinear system are not locally controllable. For a class of three dimensional, homogeneous, locally controllable, systems, we determine at which resonance values they can be stabilized.
For uncontrolled systems the interplay between resonance and instability is well illustrated by the Tacoma Narrows bridge. This paper deals with the problem of deter-mining at which resonance values an STLC nonlinear control system can be stabilized. This is answered for a class of three dimensional systems.
For ε > 0 a dilation is a mapping δεr : Rn → Rn, δεr(x) = (εr1
x1, . . . εrnxn), with
r1, . . . , rnpositive integers. A real valued function f is said to be homogeneous of degree
k with respect to a dilation δr
ε if f (δεrx) = εkf (x). We let Hrk(δεr), or just Hrk if the
dilation is understood, denote continuous functions homogeneous of degree k. Note that Hr
k need not be finite dimensional, e.g. we will consider controls u ∈ Hkr and for,
say, r = (1, 3), u(x) = (x3
1+x2)1/3is homogeneous of degree one. The notation Hr,l(δεr)
or Hr,l when the dilation is obvious, will be used to denote analytic vector fields, homogeneous of degree l with respect to δrε. A vector field X ∈ Hr,k if for all smooth f ∈ Hrl, Xf ∈ Hrl−k, l ≥ k. In local coordinates, if X(x) =Pn
i=1ai(x)∂/∂xi then
X is homogeneous of degree k if ai is analytic and in Hrri−k, i = 1, . . . , n. H1,l will
denote analytic vector fields, homogeneous of degree l with respect to the standard dilation δ1
ε having r1, . . . , rn = 1. For fixed r, l, Hr,l is a finite dimensional vector
space.
In what follows we examine four three-dimensional STLC control systems. Each contains a parameter k > 0: each is homogeneous with respect to the dilation δεrx = (εx1, ε3x2, ε9x3),and each has an uncontrollable linearization and unstable uncontrolled
dynamics. The problem is to determine values of k for which there exists a continuous, asymptotically stabilizing, state feedback control u which is homogeneous of degree one with respect to δr
ε i.e. u ∈ Hr1.
extension of the Poincar´e normal form, (hereafter Ancona normal form), there exists a free constant in the normal form which can be used to show that a homogeneous ASFC does not exist. If k is not a resonance value the system does transform to Ancona normal form and the existence of a continuous, homogeneous, ASFC is proved. For some such (limited values) k, Sepulchre, [SEP], gives an elegant geometrical construction based on work of Coron-Praly, [CP], and Kawski, [K3], to show existence of a homogeneous ASFC. This is extended in theorem 1 to all nonresonance values by constructing a system containing a free parameter which is diffeomorphic to the original system. The parameter can then be chosen to extend the Sepulchre construction. We also include open problems and conjectures. Some basic results for the four systems depend on two theorems, one due to Coron–Praly [CP],(discussed in Appendix A1-1), the other due to Kawski, [K2], and material found in the doctoral thesis of R. Sepulchre [Sep]. The use of the Coron–Praly theorem, as given in R. Sepulchre’s doctoral thesis, is also given in Appendix A1-1.
The four systems have the form
˙x1 = u, ˙x2= x2+ x31, ˙x3 = kx3+ P9(x1, x2) (0.1)
where P9 is a homogeneous polynomial, of degree 9, with respect to δεr. In general, we
denote by r = (r1, . . . , rn) the dilation exponents. For our above systems the dilation
is r = (1, 3, 9).
Kawski Resonance, Homogeneous Resonance, and Kawski’s theorem.[K3],[A1]. Homogeneous Resonance [A1]. Let X be an analytic vector field homogeneous of degree zero with respect to a dilation δrε. Expand X(x) = Ax+(higher order terms with respect to the standard dilation). Let λ1, · · · , λn, denote the eigenvalues of A.
relations of the form λi = n X s=1 νsλs, ri = n X s=1 νsrs
with ν1, · · · , νn non negative integers,Pns=1νs= l, with l ≥ 2. If there is no resonance
of any order, A is said to be non resonant.
REMARK 0-1. It is basic to note that changing the first equation in (0.1) to be ˙x1 = ax1+ v, which can always be done by introduction of the new control v(x) =
u(x) − ax1, does not change the existence, or lack of existence, of an ASFC. However,
the term ax1 can change the resonance values! We will always absorb terms in the first
equation which are homogeneous of degree one into the control and thereby have the first equation ˙x1 = u(x).
Theorem. (Ancona [A1]). Let X be a real analytic vector field, homogeneous of degree zero with respect to the dilation δr
ε. If the linear part A of X is non resonant
with respect to δr
ε there exists a real analytic, homogeneous, coordinate change (a
dif-feomorphism of Rn) which preserves homogeneity and transforms X to its linear part.
Remark 0-2. As we shall see, non resonance is sufficient, but not necessary, to trans-form X to its linear part, preserving homogeneity.
Let A(x) denote the vector field Ax and adA be defined by (adA,v)=[A,v], the Lie product of the vector fields. A key step in the proof of Ancona’s theorem is that if A has resonance of order l then the linear map adA : Hr,0∩ H1,l−1 → Hr,0∩ H1,l−1 is not onto, i.e. has a nontrivial kernel.
form need not be unique.
Suppose the nonlinear systems (0.1) can be transformed to “Ancona normal form”, i.e. there exists a change of variable which transforms system (0.1) to the form
˙y = Ay + uY (y), A = A(k) = 0 0 0 0 1 0 0 0 k . (0.2) Let R = 1 0 0 0 3 0 0 0 9
. The vector field Ry is called the Euler field associated with the
dilation δrε.
Kawski’s Theorem [K1] [K2]. If there exists a c ≥ 0 such that rank (A − cR) ≤ 1, then system (0.2) has Kawski resonance (for the given value of k) and system (0.2), hence also system (0.1), does not admit an asymptotically stabilizing, feedback control (hereafter ASFC) in Hr
1.
Coron-Rosier Lemma [C1]. Assume the n dimensional system (i) ˙x = F (x) is homogeneous with respect to a dilation δr
ε, r = (r1, . . . , rn). Let R be the diagonal
matrix R=diag (r1, . . . , rn)and assume the null solution of (i) is asymptotically stable.
Then:
(a) For m ≥ rn+ 1, system (i) admits a Lyapunov function V(x) which is homogeneous
of degree m with respect to δεr and is C1 on Rn− {0}.
(b) Any homogeneous Lyapunov function V for system (i) is also a Lyapunov function for ˙x = F (x) − cRx, c > 0.
(c) The null solution of ˙x = F (x) − cRx is also asymptotically stable for any c ≥ 0. In section 1 we will present four examples, each of the form (0.1) and homogeneous with respect to the dilation δεrx = (εx1, ε3x2, ε9x3) and STLC at the origin. This means
the origin as t → ∞) is satisfied.
From (0.1) we see each example will have the same linear part with matrix A having eigenvalues λ1 = 0, λ2 = 1, λ3 = k and dilation exponents r1 = 1, r2 = 3, r3 = 9.
Homogeneous resonance occurs at k = 3, 2, 1, 0. Indeed:
λ3 = 3 = 3λ2+ 0λ1, r3 = 9 = 3r2+ 0r1
showing k=3 gives resonance of order 3.
λ3 = 2 = 2λ2+ 3λ1, r3 = 9 = 2r2+ 3r1
showing k=2 gives resonance of order 5.
λ3 = 1 = 1λ2+ 6λ1, r3 = 9 = 1r2+ 6r1
showing k=1 gives resonance of order 7.
λ3 = 0 = 0λ2+ 9λ1, r3 = 9 = 0r2+ 9r1
showing k=0 gives resonance of order 9.
We will be interested only in the case k > 0 for which it easily follows that, with the examples in their initial form, the following necessary condition is satisfied.
Brockett necessary condition (iii), [B]. A necessary condition that system (0.1) admit a continuous ASFC is that
{(u, x2+ x31, kx3+ P9(x1, x2)) : x ∈ Rn, u ∈ R1}
covers a neighborhood of the origin in R3.
Hereafter we consider k > 0. All examples have homogeneous resonance at k = 1, 2, 3.
Transforming system (0.1) to Ancona normal Form. The general transform of (0.1) to Ancona normal form proceeds as follows. Let
y1 = x1, y2 = x2+ x13, y3 = kx3+ f (x1, x2) (0.3)
or y = φ(x), where f is a δr
ε homogeneous polynomial of degree 9. Specifically, for our
examples, we choose
f (x1, x2) = ax32+ bx31x22+ cx61x2+ ex91. (0.4)
It is useful to leave the transformation of system (0-1) in the form ˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = k(kx3+ P9(x1, x2)) + fx2(x2+ x
3
1) + fx1u.
Thus to obtain Ancona normal form we need
kP9(x1, x2) + (x2+ x31)fx2(x1, x2) = kf (x1, x2) (0.5)
which yields simple linear equations to determine the constants a, b, c, e by equating like powers of x1, x2. If (0.5) can be satisfied the system (0.1) transforms to
˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = ky3+ fx1(φ
−1(y))u.
At resonance values k, the transform may not exist, or if it exists, it is not unique, i.e. the constants a, b, c, e are not uniquely determined.
THEOREM 1. (A) The systems (1.1.1),(1.2.1),(1.3.1), (1.4.1), below, do not admit an ASFC in H1r at a resonance value k for which they can be transformed to Ancona normal form.
SECTION 1. THE FOUR EXAMPLES & THE ROLE OF RESONANCE. 1.1. Example 1. (Kawski)
˙x1 = u, ˙x2 = x2+ x31, ˙x3 = kx3+ x91. (1.1.1)
For any k 6= 0 this can be transformed to the Ancona normal form, i.e.
˙y1 = u, ˙y2 = y2+ 3y21u, ˙y3 = ky3+ 9y81u. (1.1.2).
By using the Coron-Praly theorem to eliminate the integrator and the construction, as given in the appendix, Theorem A-1, it is shown in [Sep, §6.2.1]:
(1-A) System (1.1.1) has an ASFC in H1(δrε) if k < 1 or k > 3.
(1-B) From (1.1.2) and Kawski’s theorem, there is no ASFC in H1(δεr) if k = 3. (Using
the Coron-Rosier lemma, subtract (1/3)Ry from the right side of (1.1.2) and the resulting system is easily shown to not satisfy the Brockett necessary condition (iii). The Brockett necessary condition (iii) is a vector field index condition, hence invariant under a coordinate change, such as transforming to Ancona normal form. The index can, however, be changed by subtracting cRy, c > 0.)
(1-C) For k ∈ [1, 3) it was unknown if there exists an AFSC in Hr
1. A consequence of
theorem 1 is that there is no such for k = 1, 2 but there is such for k ∈ (1, 2) and k ∈ (2, 3).
For the resonance value k = 1, of order 7, system (1.1.1) admits the Ancona normal form
˙y1 = u, ˙y2 = y2+ 3y21u, ˙y3 = y3+ (6cy51y2+ 3cy18+ 9y18)u. (1.1.3)
For the resonance value k = 2, system (1-1) admits the Ancona normal form
˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = 2y3+ (3cy12y22+ cby15y2+ 9y18)u. (1.1.4)
For resonance value k = 3 the Ancona normal form is
˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = 3y3+ (9cy12y22+ 9y18)u. (1.1.5)
One can get a reduced order, related, system to (1.1.1) as follows. Multiply the first equation by 3x2
1, define v = 3x21u, y1 = x31, y2 = x2, y3 = x3 getting the
system
˙y1 = v, ˙y2 = y2+ y1, ˙y3= ky3+ y13. (1.1.1a)
This system is homogeneous with respect to the dilation δεr having r = (1, 1, 3), again k = 1, 2, 3 are resonance values. The Sepulchre analysis of system (1.1.1a) is exactly the same as that for system (1.1.1) and the known results (1-A), (1-B), (1-C), above, hold for system (1.1.1a) also. One should note that knowing an ASFC v for system (1.1.1a) does not mean one can recover such (or that such even exists) for system (1.1.1). System (1.1.1a) is often referred to as the dynamic extension of the planar system
˙x1 = x1+ u, ˙x2 = kx2+ u3. (1.1.1b)
1.2. Example 2.
˙x1 = u, ˙x2 = x2+ x31, ˙x3 = kx3+ x61x2. (1.2.1)
For k 6= 1, this can be transformed to the Ancona normal form, ˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = ky3+
6ky5 1y2 (k − 1) + 3(3 − 2k) (k − 1) y 8 1 u. (1.2.2)
By eliminating the integrator and using the Sepulchre approach one can show (see Appendix A2-1):
(2-A) System (1.2.1) has an ASFC in H1r if 0 < k < 3/2.
(2-B) From (1.2.2) and the Kawski theorem, there is no ASFC in Hr1 if k = 3.
(2-C) For k ∈ [3/2, 3) or k > 3 it was unknown if there exists an ASFC in Hr1. Theorem 1 shows there is no such for k = 2 while for k ∈ [3/2, 2) ∪ (2, 3) and k > 3 such an ASFC does exist.
For the resonance value k = 2 system (1.2.1) admits the Ancona normal form ˙y1 = u, ˙y2 = y2+ 3y12u, ˙y3 = 2y3+ [c(3y12y22+ 6y15y2) + 12y15y2 − 3y18]u. (1.2.3)
Since Kawski’s theorem gives the result for k = 3 we will not give the general Ancona normal form for that resonance value.
The reduced order system associated with system (1.2.1) is
˙y1 = u, ˙y2 = y2+ y1, ˙y3 = ky3+ y12y2. (1.2.1a).
This system is homogeneous with respect to the dilation δrε having r = (1, 1, 3). Again, the Sepulchre theorem analysis for (1.2.1a) is exactly the same as that for (1.2.1) and the results (2-A), (2-B), (2-C) above are valid for system (1.2.1a).
System (1.2.1a) is the dynamic extension of the planar system
Sepulchre, [Sep, theorem 6.4], shows that for k = 3 system (1.2.1b) admits a continu-ous ASFC, but not a continucontinu-ous, homogenecontinu-ous ASFC. Indeed, if it had a continucontinu-ous, homogeneous, ASFC it would imply the same for system (1.2.1a). But subtracting Ry from system (1.2.1a) yields a system which fails the Brockett necessary condition (iii) and hence, by the Coron-Rosier lemma, system (1.2.1a) does not have an ASFC in Hr 1
for k = 3.
1.3 Example 3.
˙x1 = u, ˙x2 = x2+ x31, ˙x3 = kx3+ x31x22. (1.3.1)
For k 6= 1, 2, this can be transformed to Ancona normal form,
˙y1 = u, ˙y2 = y2+ 3y21u, (1.3.2) ˙y3 = ky3+ h3ky2 1(y2−y 3 1) 2 (k−2) + 6y8 1(3−2k)+12y 5 1y2 (k−1)(k−2) i u.
Again, by eliminating the integrator and using the Sepulchre approach on (1.3.1) one can show
(3-A) System (1.3.1) has an ASFC in Hr
1 if k 6= 3.
(3-B) From (1.3.2) and Kawski’s theorem, there is no ASFC in Hr
1 for k = 3.
The reduced order system associated with system (1.1.3) is
˙x1 = u, ˙x2 = x1+ x2, ˙x3 = kx3+ x1x22. (1.3.1a)
and properties (3-A), (3-B) apply to system (1.3.1a) also. Example 4. (Kawski, [K3].)
˙x1 = u, ˙x2 = x2+ x31, ˙x3 = kx3+ x32. (1.4.1)
˙y1 = u, ˙y2 = y2+ 3y21u, (1.4.2) ˙y3 = ky3+ h 9ky2 1y 2 2 (k−3)(k−2) + 18y5 1y2 (k−2)(k−1) − 9y8 1 (k−1) i u.
As shown by Kawski, and also easily shown by the Sepulchre analysis, system (1.4.1) has an ASFC in Hr
1 for all k > 0.
1.5 THE ROLE OF HOMOGENEOUS RESONANCE.
While we will deal, here, with theorem 1 which is specific to the four examples, hopefully the proof extends to the more general result which, at this point we leave as Conjecture. Consider the analytic, affine, homogeneous, n-dimensional system
˙x = X(x) + uY (x). (1.5.1)
where X is homogeneous of degree zero with respect to some dilation δεr and Y is homogeneous of degree one. Expand X = Ax +(higher order terms) and assume the eigenvalues of A are non negative. If A has homogeneous resonance with respect to δr ε
but system (1.5.1) can still be transformed to Ancona normal form, then system (1.5.1) does not admit an ASFC in Hr
1.
Remark 1.5.1. If A is allowed to have negative eigenvalues the above theorem would not be true. See example A-2 of the appendix.
adA. Therefore it can have effect in yielding an ASFC when the system is in the form ˙x = Ax + f (x) + uY (x) but the goal of the Ancona form is to remove f .
The Ancona normal form of example 3 has singularities at k = 1, 2 but not at k = 3. Again, in normal form, the control has huge effect near k = 1, 2 but not near 3. Here there is an ASFC in Hr1 except for k = 3. The Ancona normal form of example 2 has a singularity only at k = 1 but not at the other two resonance values k = 2, 3. Here there is an ASFC in Hr1 for all k > 0, except for k = 2, 3.
In example 1, there are no singularities in the Ancona normal form, i.e. no strong control effects near or at the resonance values k = 1, 2, 3. We know an ASFC in Hr 1
exists for k < 1 or k > 3. There is no such at k = 1, 2, 3.
1.6 The proof of THEOREM 1. We will provide, here, only a proof of the statement A, with k = 1, 2 of THEOREM 1 relative to the system (1.1.1), the case of system (1.2.1) being entirely similar. We next show the arguments for statement B for system (1.1.1) with k = 1.5 and for system (1.2.1) for k = 3/2 and k = 4. The remaining cases can be shown by similar constructions and therefore are not included. In the case of system (1.3.1) the conclusions follow from Kawski’s Theorem 1.2 while system (1.4.1) cannot be transformed to Ancona normal form for any resonance value and all results are known.
A. We shall begin by showing that there is no ASFC in Hr
1 for the system (1.1.1) for
resonance value k = 1. For the resonance value k = 1, which is of order 7, the Ancona normal form (1.1.3) can be obtained from the general form (1.1.2) by a coordinate transformation. Specifically:
z1 = y1, z2 = y2, z3 = y3+ cy61y2 (1.6.1)
where the term cy16y2 results from the resonance of order 7. Assume (1.1.1) has a
which we denote by u0(y). Therefore, for every c, system (1.1.3) has an ASFC denoted uc. From the variable change (1.6.1), for any c
uc(y1, y2, y3) = u0(y1, y2, y3− cy16y2) (1.6.2).
Thus to show (1.1.1) does not have an ASFC for k = 1 it suffices to show there exists a value c∗ for which system (1.1.3) does not admit an ASFC. For k = 1 and
notational ease denote the system (1.1.2) as ˙y = f0(y, u0(y)) and system (1.1.3) as ˙y = fc(y, uc(y)) where u0, uc are related by (1.6.2).
Construction of c∗ for k=1. Following Kawski, [K3], let ν(x) =P r
ixi∂/∂xidenote
the Euler vector field (which generates homogeneous rays). Let m = 2r1r2r3. Define
the (smooth) δεr unit two sphere Sr2 = {x ∈ R3 :P xm/ri
i = 1}. Let π : (R3− 0) → Sr2
be defined by π(x) is that point y ∈ Sr2where the homogeneous ray through x intersects Sr2. Since π(δerctx) = π(x) for all t, evaluating a t-derivative of the previous at 0 one
finds π∗(ν(x)) = 0. Also, if F (x) = P ai(x)∂/∂xiis a smooth vector field homogeneous
of degree zero, one can compute that the Lie product [F, ν] = 0 and π∗F is well defined,
i.e. for y ∈ S2
r, π∗F (δercty) = π∗F (y). If y is such that π∗F (y) = 0 then F (y) is a
scalar multiple of ν(y) and this holds along the homogeneous ray through y hence the solution of ˙x = F (x), x(0) = y lies on the homogeneous ray through y. We proceed formally for a moment. Choose a value c0 ; define π∗fc0(z, uc0(z)) as homogeneous
projection of fc0
onto the unit sphere Sr2. Then π∗fc0 must have a zero on Sr2 (it may
have several), say at zc0
= (zc0
1 , zc
0
2 , zc
0
3 ) and the solution of (1.1.3) through zc0 then
lies on the homogeneous ray through zc0. Specifically, this ray has the form
z1(t) = zc10et, z2(t) = z2c0e3t, z3(t) = z3c0e9t.
(a) If zc0
1 = 0, the solution of (1.1.3) lies in the plane z1 = 0 and the third component
(b) For zc0
1 6= 0, the first two components of the solution y(t) of (1.1.3) through zc0 lie
on the curve y2/y13 = zc 0 2 /(zc 0 1 )3.
For notational ease let e0 = zc0
2 /(zc
0
1 )3 and (for later) ei = zc2i/(zc1i)3. Then y2(t) −
e0y3
1(t) ≡ 0. Differentiating this and combining with the second equation of (1.1.3)
yields
y2(t) = 3y12(t)uc0(e0− 1). (i)
If e0 6= 1, from (i), 3y21(t)uc0 = y
2(t)/(e0 − 1) and the second component of (1.1.3)
becomes
˙y2+ 3y12(t)uc
0
= y2(t)(1 + (1/(e0− 1))) = e0y2(t)/(e0− 1). (ii)
Notice that e0/(e0− 1) > 0 iff e0 ∈ (−∞, 0) ∪ (1, ∞). Therefore, in the case zc0
1 6= 0, it
follows that the second component of the solution y(t) of (1.1.3) through zc0 is unstable
whenever e0 ∈ (−∞, 0) ∪ (1, ∞).
(c) We proceed, by induction, to prove the existence of a constant c∗ and equilibrium
point z∗ of π
∗fc∗ so that the solution of (1.1.3) through z∗ is unstable.
Step 1: Fix c0 ∈ R, let zc0 = (z1c0, z2zc0, zc30) be a zero of π∗fc0. If either z1c0 = 0 or
e0 ∈ (−∞, 0) ∪ (1, ∞) we have instability by (a), (b) and we are done. Otherwise we have zc0
1 6= 0, e0 ∈ [0, 1] and we may define
c1 = −3/(1 + 2e0) ∈ [−3, 1]. (iii)
Notice that c1 is defined to make the coefficient
[c1(2z2c0 + (zc
0
1 )3+ 3(zc
0
1 )3]
of u equal to zero in the third equation of (1-3). Now let zc1 be a zero of π
∗fc1, c2
be defined by (iii) with e0 replaced by e1, etc. Inductively this defines a sequence of constants {ci}, with ci ∈ [−3, −1] and points zci−1 ∈ Sr2. If ever z
ci−1
ei−1 ∈ (−∞, 0) ∪ (1, ∞) we may stop and the solution of (1-3) through zci−1 will be
unstable. If this does not occur, we obtain sequences {ci}, {zci−1} having values in the
compact sets [-3,-1], S2
r respectively, hence there exist subsequences which converge to
c∗ ∈ [−3, −1], z∗ ∈ Sr2.
Note. We need not worry about convergence of {uci} since uc∗ is given in terms of the
assumed stabilizing control u0 by (1.6.2).
Then π∗fc∗(z∗, uc∗) = 0, c∗ = −3(z∗1)3/((z1∗)3+ 2z2∗) and the third component of the
solution of (1.1.3) through z∗ satisfies ˙y
3 = y3 and is unstable.
Construction of c∗ for Example 1, k=2. With the notation of the previous case
for k = 1, choose a real c0, let zc0 ∈ Sr2 be a zero of π∗fc0 where now fc0 is given by
(1.1.4) with c replaced by c0. Again, if z1c0 = 0 the solution of (1.1.4) lies in the plane
z1 = 0 and the third component of (1.1.4) is ˙y3 = 2y3 and unstable.
If z1 6= 0 and e0 = z2c0/(zc
0
1 )3 ∈ (−∞, 0) ∪ (1, ∞), the second component of (1.1.4)
reduces to ˙y2 = e0y2/(e0− 1) with e0/(e0− 1) > 0 and we have instability.
Now to make the coefficient of u equal to zero in the third equation of (1.1.4), formally we need c1 = −3(z1c0)6/(zc 0 2 + 2(zc 0 1 )3)zc 0 2 or if zc0 1 6= 0, c1 = −3(z1c0)3/(e0 + 2)zc 0 2 . (iv) If zc0
2 = 0 the homogeneous ray through zc0 on which the solution of (1.1.4) through
zc0 lies has second component which satisfies ˙y
2 = 3y12u. But from the first equation,
3y2
1(t) ˙y1 = d/dt(y13(t)) = 3y12u so subtracting shows d/dt(y2(t) − y13(t)) = 0, y2(t) −
y13(t) = zc0
2 − (zc
0
1 )3 = (zc
0
1 )3. But y2(t) ≡ 0 hence y1(t) = −z1c0 is a nonzero constant
and this shows instability. In summary, if either zc0
1 = 0 or zc
0
2 = 0 or e0 ∈ (−∞, 0) ∪ (1, ∞) we may stop since
there would be instability. Thus assumezc0
1 6= 0, zc
0
(iv) will define c1 ∈ [−3/2, 3/2] and one can continue, inductively, to obtain sequences
{ci}, {zci} which have convergent subsequences converging to a real c∗ ∈ [−3/2, 3/2]
and z∗ ∈ S2
r with the third component of the solution of (1.1.4) through z∗ being
˙y3 = y3 and hence unstable.
The details for instability, with k = 2 in Example 2 are similar and will be omitted.
B. We begin with seeking the most general system of the form
˙x1 = u, ˙x2 = x2+ x31, ˙x3 = kx3+ P9(x1, x2) (1.6.3)
with k > 0, which has (1.1.2) as its Ancona normal form. For such a system, since both it and (1.1.1) smoothly transform to the same system, they can be smoothly transformed to each other and hence one has a homogeneous ASFC if and only if the other also does.
In (1.6.3) choose
P9 = αx91+ βx32+ γx22x31+ δx2x61.
Let f (x1, x2) be given by (0-4), the transform y = φ(x) be given by (0-3) and explicitly,
fx1(φ
−1(y)) = 3by2
1y22− 6by15y2+ 3by18+ 6cy15y2− 6cy18+ 9ey81.
We use (0.5) to attempt to transform (1.6.3) into (1.1.2). Equating like powers, using (0.5), requires
a = βk/(k − 3), b = γk/(k − 2) + 3βk/(k − 2)(k − 3), (iii)
c = 2γk/((k − 1)(k − 2)) + 6βk/((k − 1)(k − 2)(k − 3)) + δk/(k − 1), e = (kα + c)/k = α + (c/k).
We assume, hereafter, that k 6= 1, 2, 3, i.e. the system we obtain will transform to system (1.1.2) except at the resonance values. Substituting these into fx1(φ
which we need to be equal to 9y18 in order to have the Ancona transform of (1.6.3) be (1.1.2) requires b = 0 or 2γk(k − 3) + 6βk = 0, c = 0 or δk(k − 2)(k − 3) = 0 which requires δ = 0, e = 1 or α = 1. Thus, from b = 0, γ = −3β/(k − 3) with β arbitrary. Therefore the most general system of the form (1.6.3) which has Ancona transform system (1.1.2) is
˙x1 = u, ˙x2 = x2+ x31, (1.6.4)
˙x3 = kx3+ x91+ βx23 − 3(β/(k − 3))x31x22, k 6= 1, 2, 3.
System (1.6.4) has the advantage of the free constant β. We now use the generalized Sepulchre method of Appendix A to study system (1.6.4). Sepulchre [Sep] showed that for 0 < k < 1 and for 3 < k the system (1.1.1) admits a homogeneous ASFC. we next use THEOREM A-1 (Generalized Sepulchre construction) to show that for 1 < k < 2 and for 2 < k < 3 there does exist a homogeneous ASFC.
Following the construction in appendix A, let
P9(v1/3, θ) = v3+ βθ3− ((3β/(k − 3))vθ2.
In short, THEOREM A-1 states:
For P9 depending on both x1 and x2 a sufficient condition that equation (1.6.4) has a
homogeneous ASFC is that there exists a line, having nonzero slope, in the (θ, v) plane which intersects the set
Z = {(θ, v) : −βθ4 + (3β/(k − 3))vθ3− (v3+ k − 3)θ + 3v = 0} (1.6.5)
only in points (θ∗, v∗) with either v∗/θ∗ < −1 or
P9((v∗)1/3, θ∗) < −k, if θ∗ 6= 0.
To illustrate the idea to show such a line is possible consider:
curves of Z. For notational ease let x = θ, y = v in Z. Use ContourPlot (and be sure to leave spaces between x3 and y and also x and y3) specifically:
ContourPlot[0.5x4+ x3y − xy3+ 1.5x + 3y, {x, −10, 10}, {y, −10, 10}].
Move the cursor around to find the (obvious) zero level curves. You now want a line, say y = c1x + c2, which intersects these only in points (x∗, y∗) with y∗/x∗ < −1 or
P9((y∗)1/3, x∗) < −1.5, x∗ 6= 0. See Fig. 1.
-10 -5 0 5 10 -10 -5 0 5 10 Example 1, k=1.5, beta=-0.5 Fig 1, system (1.6.4), (k = 3/2, β = −0.5)
Substituting y = c1x+c2 one can use MATHEMATICA to seek these intersections,
NSolve[0.5x4+ 1x3(c1x + c2) − x(c1x + c2)3+ 1.5x + 3(c1x + c2) = 0, x]
For general c1, this is a quartic and will have four roots (x∗, y∗) with at least
one not satisfying THEOREM A.1. HOWEVER, if c1 satisfies the cubic equation
−c3
1+ 1c1+ 0.5 = 0 then instead of a quartic in the NSolve one gets a cubic, i.e. the x4
terms vanish. The roots of this cubic are c1 = 1.19149 and then two complex numbers.
(It is instructional to roughly sketch the lines corresponding to c1 with various negative
values of c2 in the MATHEMATICA output of the ContourPlot and see (roughly) where
the lines intersect the zero level curves.) For c2 = −3 there are three real crossings,
with x values x=0.47917, x=1.16964, x=1.41774, and all are ”bad” crossings, i.e. have y∗/x∗ > −1. For c
2 = −8 the line y = 1.19149x − 8 crosses the zero curve of the
ContourPlot at x∗ = 0.047416 with corresponding y∗ = −7.9435 (the other crossings
have complex values, except for x = 3.78063(10)6 which is a numerical round off root
since with an exact value of c1 the cubic in the ContourPlot equation will only have
three roots.) Here y∗/x∗ < −1. Hence (1.6.4) has a homogeneous ASFC for k=1.5
and β = −0.5 and (1.1.1) also has a homogeneous ASFC for k=1.5. (It is instructional to set β = 0 in the ContourPlot to see how the zero curves (obtained by Sepulchre [Sep])show an inability to find a line crossing them only at points satisfying Theorem A-1,and how these zero curves compare with the case β = −0.5.) With β = −0.5 and any k ∈ (1, 2) or k ∈ (2, 3) the above construction leads to a line satisfying THEOREM A.2.
For Example 1 with k ∈ (2, 3), say k = 5/2, choose β = −0.2 in (1.6.5) and again choose a line with slope so that the coefficient of x4 is zero. Here this requires c
1 = −0.163054.
Choose the line y = −0.163054x−10 and Mathematica shows this line crosses the curve Z = 0 with real root x = 0.0299554, two complex roots (and the numerical error root x = 8.682(10)7 ). The crossing (x∗ = 0.0299554, y∗= 10.0048) shows Example 1 has a
We next turn to system (1.2.1) of Example 2 and its Ancona transform system (1.2.2). The direct approach using the Generalized Sepulchre THEOREM A-1 shows the existence of a homogeneous ASFC if 0 < k < 3/2. Details showing this direct approach will not work at k = 3/2 or k > 3 are given in appendix A2-1. We next briefly outline how to show such a control exists for k ∈ [3/2, 2) ∪ (2, 3) or k > 3. Again, we seek the most general system of the form (1.6.3) which has system (1.2.2) as its Ancona transform. Choose f as in (0-4), form (0-5), equate like powers to determine the constants, and one finds: For k 6= 1, 2, 3:
The most general system of the form (1.6.3) which has Ancona transform (1.2.2) is ˙x1 = u, ˙x2= x2+ x31, ˙x3 = kx3+ (1.6.6)
(βk/(k − 3))x91+ βx32− (3β/(k − 3))x22x31+ x2x61, k 6= 1, 2, 3.
Note that if β = 0 we get back system (2-1). For application of THEOREM A-1, here P9(x1, x2) = (βk/(k − 3))x91+ βx32− (3β/(k − 3))x22x31+ x2x61.
Again, let x = θ, y = v1/3 in P
9(v1/3, θ) and use MATHEMATICA to graph the
zero curves of Z as given in the appendix by (A-6), with various values of k and β as illustrations. This leads to
ContourPlot[(k−3)−βk xy3−βx4+(k−3)3β x3y −x2y2+(3−k)x+3y, {x, −10, 10}, {y, −10, 10}]
In APPENDIX A2-2 it is shown that the direct use ot THEOREM A-1 on system (1.2.1) for k > 3 will not give the existence of an ASFC in Hr
1 but will for k < 3/2.
We next illustrate the use of the modified system (1.6.6) to show the existence of a homogeneous ASFC for k = 3/2 by choosing β = −0.2. (This construction works simililarly for k ∈ (3/2, 2) ∪ (2, 3).) We, again, use MATHEMATICA to plot level curves of Z. Here
-10 -5 0 5 10 -10 -5 0 5 10 Example 2, k=32, beta=-0.2
Fig.2, Ex.2 , system (1.6.6),(k = 3/2, β = −0.2)
From the graph, a line such as y + 7 = (−5.4/2)(x − 1) i.e. y = −2.7x − 4.5 seems reasonable. Solving (via MATHEMATICA), this line has two real intersections with the zero curve of Z, specifically x∗ = 0.794305, with y∗ = −6.205695 and x∗ = 1.48284 with
y∗ = −8.2071. See fig.2. Thus for each y∗/x∗ < −1 and there exists a homogeneous
ASFC at k = 3/2.
The construction for the existence of a homogeneous ASFC for k ∈ (3/2, 2) and k ∈ (2, 3) is similar.
-4 -2 0 2 4 -10 -5 0 5 10 Example 2, k=4, beta=-0.1
Fig. 3, Ex. 2, system (1.6.6), (k = 4, β = −0.1)
all k > 3.
With k = 4, choose β = −0.1 and:
ContourPlot [−.4xy3− 0.1x4+ 0.3x3y − x2y2− x + 3y, {x, −10, 10}, {y, −10, 10}].
A line, for example, y = −4.5 has two real crossing of the zero curve of Z at x∗ =
0.582467 and x∗ = 1.02471, both with y∗ = −4.5, hence both crossings satisfy Theorem
A-1 and system 2-1 has a homogeneous ASFC for k=4. Note that here P9(0, x2) = βx32
so the horizontal line is permissible. See fig. 3.
A3-1 by direct use of Theorem A-1.
APPENDIX. The object of study in this section will be a reasonably general, three dimensional, homogeneous, affine control system for which the uncontrolled dynamics are unstable and the linearization is uncontrollable. We begin with a general setting for Sepulchre’s methods, [Sep]. Specifically, we study the system
˙x1 = u, ˙x2 = k2x2+ x2m+11 , ˙x3 = k3x3+ P2l+1(x1, x2) (A.1)
where 0 ≤ m < l are integers and P2l+1(x1, x2) is a homogeneous polynomial of degree
(2l + 1) with respect to the dilation δr
ε(x) = (εx1, ε2m+1x2, ε2l+1x3),i.e.
P2l+1(εx1, ε2m+1x2) = ε2l+1P2l+1(x1, x2). Most cases of interest will have k2 > 0, k3 >
0, however k2 < 0 is useful for example A-2. If u is continuous and homogeneous of
degree one with respect to δεr we again write u ∈ Hr1. For such a control u, the vector field on the right side of (A.1) will be homogeneous of degree zero. We will assume STLC (which in some cases restricts the values of k ) and seek an ASFC u ∈ Hr
1 for
(A.1). One should note that system (A.1) with k2> 0, k3 > 0 does satisfy the Brockett
necessary condition (iii). By choosing various polynomials P2l+1many examples can be
examined. The method used in the analysis of these examples is a generalization of ho-mogeneous dimensional reduction introduced by Kawski, [K1], [K2], and an abstraction of the procedure used by Sepulchre [Sep].
The analysis of (A.1) begins by invoking the Coron-Praly theorem, [CP], on the elimination of the integrator. Let w = x2m+11 or x1 = w1/(2m+1) and consider the two
dimensional system
˙x2 = k2x2+ w, ˙x3 = k3x3+ P2l+1(w1/(2m+1), x2). (A.2)
Let γεr(x2, x3) = (ε2m+1x2, ε2l+1x3). Then if w is homogeneous of order 2m + 1 with
Coron-Praly theorem [CP]. If the homogeneous system (A.2) admits an ASFC in Hγ2m+1 then the homogeneous system (A.1) admits an ASFC in Hr
1.
Remark A-1. The proof is not constructive, i.e. knowing the ASFC for (A-2) does not yield an ASFC for (A-1).
The Coron-Praly theorem reduces the existence problem from dimension three to dimension two. The reduction to dimension one is a homogeneity reduction, as given by Kawski [K1], [K2]. Let
α = (2m + 1)/(2l + 1), θ = x2/xα3, v = w/xα3. (A.3)
Notice that θ ∈ (−∞, ∞) parameterizes homogeneous curves, i.e. x2 = θxα3 is a
homo-geneous curve in the plane. The transformation (A.3) may be viewed as homohomo-geneous projection from the plane to the projective line P1. Since system (A.2) is homogeneous, the vector field induced on P1 by the transformation (A.3) is well defined and easily
computed. Indeed
˙θ = k2θ + v − αk3θ − αθ[(1/x3)P2l+1(w1/(2m+1), x2)].
For ease of exposition we note that for w ∈ Hγ2m+1, w1/(2m+1) has weight one while
x2 has weight (2m + 1) relative to γεr. Thus a possible term in P2l+1 would be
x2l/(2m+1)2 w1/(2m+1). In this case, (1/x3)x2l/(2m+1)2 w1/(2m+1)= v1/(2m+1)θ2l/(2m+1).
In general
(1/x3)P2l+1(w1/(2m+1), x2) = P2l+1(v1/(2m+1), θ). (A.4)
Thus the induced equation on P1 is
If v∗(θ) is a linear function of θ, say v∗(θ) = c
1θ + c2, the control function w which
corresponds to v∗ via (A.3) is w(x
2, x3) = xα3(c1θ + c2) = c1x2 + c2xα3 and this is a
control in Hγ2m+1. If (θ, v∗(θ)) is a zero of the right side of (A.5) with v replaced by
v∗(θ) then θ corresponds to an invariant homogeneous curve of (A.2). Should values of
θ exist such that (θ, v∗(θ)) is a zero of the right side of (A.5), these will comprise the
ω -limit set of (A.5) with v = v∗(θ). Kawski’s theorem [K2], shows that if the origin
of the restriction of system (A.2) to invariant, homogeneous curves corresponding to such values θ in the ω-limit set of (A.5) is asymptotically stable, then system (A-2) is asymptotically stable. (For an odd system such as (A.1), on can replace S1 by P1 in
Kawski’s theorem.)
Observe that the transformation (A.3) is only defined for x3 6= 0 hence the study
of (A.5) may exclude part of the ω-limit set of system (A.2). Specifically, we also have to check if the point at infinity of P1 is an equilibrium point for (A-5) or equiv-alently that the ray through x3 = 0 on S1 in the (x2, x3) space is an invariant
curve of (A.2). This will be the case if w(x2, x3) = c1x2 + c2xα3 in (A.2) and if
P2l+1((c1x2+ c2xα3)1/(2m+1), x2) vanishes at x3 = 0, i.e. if P2l+1((c1x2)1/(2m+1), x2) =
0. Furthermore, in this case (A-2) shows ˙x2 = (k2+ c1)x2 on this ray. Hence there is
stability if c1 < −k2, instability if c1 ≥ −k2. (Certainly if c1 = 0 and P2l+1(0, x2) = 0
then an unstable solution of (A.2) exists on the ray through the point x3 = 0 on S1 in
the (x2, x3) space but this is not the most general case.) Let
Z = {(θ, v) : k2θ + v − αk3θ − αθP2l+1(v1/(2m+1), θ) = 0}. (A.6)
We seek a line v∗(θ) = c
1θ + c2 which intersects the zero set of Z in points (θ, v∗(θ))
that x3˙x3 < 0 or x2˙x2 < 0. Computing, using (A.4),
x3˙x3 = x23[k3+ P2l+1(v1/(2m+1), θ)], (A.7)
x2˙x2 = x22(k2+ v/θ). (A.8)
Thus we have contraction at a point (θ, v) ∈ Z if
P2l+1(v1/(2m+1), θ) < −k3 or v/θ < −k2. (A.9)
We summarize this construction as
THEOREM A-1, Generalized Sepulchre construction, [Sep]. Consider the system
˙x1 = u, ˙x2 = k2x2+ x2m+11 , ˙x3 = k3x3+ P2l+1(x1, x2) (A.1)
where 0 ≤ m < l are integers and P2l+1 is a homogeneous polynomial of degree (2l + 1)
with respect to the dilation δrε(x) = (εx1, ε2m+1x2, ε2l+1x3). Let α = (2m + 1)/(2l + 1).
A sufficient condition that system (A.1) admit an ASFC in Hr1 is that there exist a line v∗(θ) = c
1θ + c2 which intersects the zero set
Z = {(θ, v) : k2θ + v − αk3θ − αθP2l+1(v1/(2m+1), θ) = 0} (A.10)
only in points (θ, v) satisfying (A.9) and if P2l+1((c1x2)1/2m+1, x2) ≡ 0 then c1 < −k2.
Remark A-2. If a line v∗(θ) = c
1θ + c2 exists which satisfies the above theorem,
w∗(x
2, x3) = c1x2+ c2xα3 will be an ASFC for (A.2) in Hγ2m+1. The construction of an
ASFC in Hr
1 for (A.1) from w∗ is, to our knowledge, a difficult and open problem.
Remark A-3. The choice of v∗(θ) as a line is convenient but by no means necessary.
Indeed, limθ→∞v∗(θ)/θ = limθ→ −∞v∗(θ)/θ suffices. Thus one could consider a
function such as v∗(θ) = (c
1θ1/3+ c2)3. An interesting (and useful for considerations of
Remark A-4. In the case (θ, v) ∈ Z, θ 6= 0, one can see from (A.6) that k2+ v/θ = α[k3+ P2l+1(v1/(2m+1), θ)].
Hence since α > 0 it follows that either the inequalities in (A.9) are both satisfied or both fail. Thus with θ 6= 0 it is sufficient to only check one.
Remark A-5. If k3 = k2/α and P2l+1(x1, x2) is not only a function of x2 i.e. P2l+1
does depend on x1, then v = 0 is part of the zero set in (A.5) and any line with
c1 6= 0)must cross v = 0. But then P2l+1(0, θ) = 0 hence the homogeneous, invariant,
curve x2 = θxα3 of (A.2) is not contracting, and there will not be a homogeneous ASFC
for (A.2). Note that in the four examples of section 1, α = 1/3 and k3 = 3k2 is precisely
the condition which yields rank (A − cR) ≤ 1, i.e. Kawski’s theorem, to be invoked when c = 1/3. However, in general, negative results for (A.2) do not imply negative results for (A.1), i.e. the converse of the Coron-Praly theorem is not necessarily true, see [SEP, thm. 5.7].
Examples of the use of Theorem A-1.
Example A-0, (easy example). Consider the system
˙x1 = u, ˙x2 = x1, ˙x3 = kx3+ x31, k > 0. (A.11)
This is example 2.1 in [H4] where a tedious computation using nonlinear regulator re-sults showed there exists a continuous ASFC. Above, letting X(x) = x1∂/∂x2+ (kx3−
x3
1)∂/∂x3, Y = ∂/∂x1 one finds the Lie brackets Y (0), [X, Y ](0) and [Y, [Y, [Y, X]]](0)
are linearly independent showing the system is odd, STLC, [Su]. To use theorem A-1, we note that the system is homogeneous with respect to the dilation δεrx = (εx1, εx2, ε3x3), i.e. m = 0, l = 1, α = 1/3, P3(x1, x2) = x31, k2 = 0, k3 =
3v/(k − v3). It thus follows that one can always construct a line v∗(θ) = c
1θ + c2
with c1, c2 > 0 such that this line is tangent to the branch, in the fourth
quad-rant, of the zero set of Z and has only this tangency point, call it (θ, v), in com-mon with Z. Since v < −k, θ > 0 in the fourth quadrant, v/θ < 0 = k2. Also,
P3(v, θ) = v3 < −k = −k3. Theorem A-1 applies to show the system (A.10) has an
ASFC in Hr1.
If we modify equation (A.11) slightly we may lose the existence of an ASFC. Specifically, consider
˙x1 = u, ˙x2 = ǫx2+ x1, ˙x3 = 3ǫx3+ x31, ǫ > 0. (A.12).
The dilation exponents are r = (0, ǫ, 3ǫ). The eigenvalues of the linear part of the system are (0, ǫ, 3ǫ) and there is homogeneous resonance. Let R be the matrix R=diag(1,1,3). The system obtained by subtracting ǫRx from (A.12) fails the Brocket condition (iii) hence has no continuous ASFC and hence by the Coron-Rosier lemma, system (A.12) does not have a continuous ASFC. This shows that a first order perturbation can destroy the existence of a continuous ASFC. Many of the listed results for the four examples in section 1 were obtained by use of theorem (A-1). The major effort, in most cases, is the graphing of the zero set Z.
APPENDIX A2-1, Analysis of system (1.2.1) for k = 3. For system (1.2.1), in theorem A-1, m = 1, l = 4, k2 = 1, k3 = k, α = 1/3, P9(v1/3, θ) = v2θ and
(θ, v) ∈ Z requires
θ2v2− 3v + θ(k − 3) = 0. (A2-i)
v
θ
Fig A2-1 (k = 3)
Any line v = c1θ + c2, c1 6= 0, c2 6= 0, crosses the line v = 0 at a point which does
not satisfy either of the conditions (A.9). A line of negative slope through the origin, e.g. say v = c1θ, c1 < −1, may cross the branch of the graph in the second quadrant
at a permissible point; its crossing at the origin will not satisfy the first of conditions (A.9) while the second condition is the indeterminate form 0/0. This condition arises from the contraction condition x2˙x2 < 0 which (when not simplified as in (A.9)) is
x2˙x2 = x22+ x2w = x22+ x2vx1/33 . (A.13)
Thus when θ = v = 0, x2˙x2 ≥ 0 , contraction does not occur, i.e. the second of
conditions (A.9) is not satisfied. Thus, as expected, for k = 3 we cannot show the existence of an ASFC, which agrees with the knowledge from Kawski’s theorem which shows there is no homogeneous ASFC.
quadratic solutions ((θ, v) 6= 0) v = 3 ±p9 − 4θ 3(k − 3) 2θ2 , θ = (3 − k) ±p(k − 3)2+ 12v3 2v2 .
(See figures A2-2,A2-3.)
θ v
Fig A2-2 (k > 3)
Any line crossing the branch in the second quadrant must also cross the branch in the first or third quadrant, or at the origin. In the first and third quadrants, v/θ > 0 and at the origin, (A.13) shows failure of x2˙x2 < 0. A line tangent to the graph at
the minimum point (2(3−k)122/31/3, −(
(k−3)2
12 )1/3) = (θ, v) having zero slope need not be
cannot conclude the existence of a homogeneous ASFC for (1.2.1) when k > 3 using the direct approach.
Case k < 3. We refer to figure A2-3.
2/3 3 2 v θ 91/3 , 1/3 (4(k−3)) 2/3 (k−3) Fig A2-3 (k < 3)
Since v2θ ≥ 0 in quadrants one and four, the first of conditions (A.9) will not
apply. For the second of conditions (A.9), the best possibility will be a line tangent to the graph at a point on the graph just below the point (4(k−3)9 )1/3, 3/2(4(k−3)
9 )2/3).
At this point v/θ = 2/3(k − 3) hence (v/θ) < −1 if k < 3/2. One concludes that for k < 3/2 system (1.2.1) does admit a continuous, homogeneous, ASFC.
A-1, m = A-1, l = 4, k2 = 1, k3 = k, α = 1/3, P9(v1/3, θ) = vθ2. Then (θ, v) ∈ Z
requires
θ + v − (k/3)θ − θ3v/3 = 0, or v = θ(k − 3)/(3 − θ3).
For k = 3 the graph consists of the lines v = 0 and θ = 31/3 and it easily follows (as
expected) that a line satisfying either of conditions (A.9) does not exist. From Kawski’s theorem, system (1.3.1) does not have a continuous ASFC at k = 3.
31/3
v
θ
k1/3
Fig A3-1 (k > 3)
v
θ
31/3 k1/3
Fig A3-2 (k < 3)
if A has even one negative eigenvalue the system may have an ASFC in Hr 1.
˙x1 = x1+ w, ˙x2 = −x2 + x13, ˙x3 = x3+ x91. (A.14)
Note that one could absorb the term x1 into the control w in the first equation, which
would change the eigenstructure, and resonance values, but not the existence of an ASFC. In order to remain in dimension three it is useful to retain the first equation as it is.
The system is homogeneous with respect to the dilation δr
ε(x) = (εx1, ε3x2, ε9x3)
and STLC at zero. The homogeneous variable change y1 = x1, y2 = −x2+x31, y3 =
x3 + x91 transforms the system to Ancona normal form ˙y1 = y1 + w, ˙y2 = −y2 +
3y12w, ˙y3 = y3 + 9y18w. The eigenvalues of the linear part are λ1 = 1, λ2 =
showing resonance. To show an ASFC in Hr1 exists we use theorem (A-1) with,
now, x1 + w replaced by u. Then k2 = −1, k3 = 1, α = 1/3, m = 1, l =
4, P9(x1, x2) = x91 in (A-1). The zero set of Z is the graph, in the (θ, v) plane, of
θ = 3v/(4+v3) and one can readily construct a line v = c
1θ +c2 which is tangent to the
branch of the graph lying in the fourth quadrant at a point (θ, v) with v/θ < −k2 = 1
and having no other points in common with Z. Theorem A-1 gives the existence of an ASFC in Hr1 for system (A.14).
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